Let us prove the statement for \(M^{\mathsf {L}}_n(z) \) .
The matrix function \(M^{\mathsf {L}}_n(z) \)  is holomorphic in \( [0,+\infty )\)  by definition, cf. (REF ).
Due to the fact that \(Z^{\mathsf {L}}_n(z)\)  has a constant jump on
\((0,1) \cup (1,+\infty )\) , cf. (REF ) and (), the matrix function \(\displaystyle \big (Z^{\mathsf {L}}_n\big )^{\prime } \)  has the same constant jump on \((0,1) \cup (1,+\infty )\) , so that the matrix \(M^{\mathsf {L}}_n(z) \)  has no jump on \((0,1) \cup (1,+\infty )\) , and it follows that
at \(z=0\)  and \(z=1\) , \(M^{\mathsf {L}}_n(z) \)  has an isolated singularity.
From (REF ) and (REF ) it holds
\(M^{\mathsf {L}}_n (z)&= \big (Z^{\mathsf {L}}_n\big )^{\prime } (z)\big ({Z^{\mathsf {L}}_n} (z) \big )^{-1}\\\nonumber & = \big (Y^{\mathsf {L}}_n\big )^{\prime } (z)\big ( {Y^{\mathsf {L}}_n} (z) \big )^{-1}+ \frac{1}{z(z-1)} Y^{\mathsf {L}}_n (z)\left[ \begin{matrix}h^{\mathsf {L}}(z) & 0_N\\0_N & -h^{\mathsf {R}} (z)\end{matrix} \right]\big ( {Y^{\mathsf {L}}_n} (z)\big )^{-1},\) 

where
\(Y^{\mathsf {L}}_n\)  is given in (REF ).
Each entry of the matrix \(Q^{\mathsf {L}}_{n} (z)\)  is the Cauchy transform of certain function, \(f\) , of type
\(f(z)=\sum _{j \in I} \varphi _{j}(z) z^{\alpha _{j}}(1-z)^{\beta _j} \log ^{p_{j}} (z) \log ^{q_{j}} (1-z),\) 

where \(\varphi _{j} (z)\)  is, for each \(j \in I\) , an entire function with \(\operatorname{Re}(\alpha _{j})\) , \(\operatorname{Re}(\beta _{j})>-1\) ,
\( p_{j}\) , \(q_j \in \mathbb {N}\) , and \(I\)  is a finite set of indices.
It's clear that
\(\lim _{z \rightarrow 0} z f(z)=0_N&&\text{and}&&\lim _{z \rightarrow 1}(1-z)f(z)=0_N.\) 

By [1]} and [2]}, we deduce that the Cauchy transform of \(f\)  have the same properties:
\( \lim _{z \rightarrow 0}z \int _0^1 \frac{f(t)}{t-z}\, \operatorname{d}t =0_N&&\text{and}&&\lim _{z \rightarrow 1}(1-z) \int _0^1 \frac{f(t)}{t-z} \, \operatorname{d}t =0_N .\) 

Now, we will prove that
\(\lim _{z \rightarrow 0}z^2 \left( \int _0^1 \frac{f(t)}{t-z}\, \operatorname{d}t\right)^\prime =0_N&& \text{and} &&\lim _{z \rightarrow 1}(1-z)^2\left( \int _0^1 \frac{f(t)}{t-z} \, \operatorname{d}t\right)^\prime =0_N.\) 

In fact,
\(& z(1-z) \left( \int _0^1 \frac{f(t)}{t-z}\, \operatorname{d}t \right)^\prime = \int _{0}^1 \frac{z(1-z) f(t)}{(t-z)^{2}} \operatorname{d} t\\& \phantom{olaolaolaol}=\int _{0}^1 \frac{(t-z)(t+z-1) f(t)}{(t-z)^{2}} \operatorname{d} t+\int _{0}^1 \frac{t(1-t) f(t)}{(t-z)^{2}} \operatorname{d} t, \\& \phantom{olaolaolaol}=\int _{0}^1 \frac{t+z-1}{t-z}f(t) \operatorname{d} t-\left.\frac{t(1-t) f(t)}{t-z}\right|_{0}^1+\int _{0}^1 \frac{\left(t(1-t) f(t)\right)^{\prime }}{t-z} \operatorname{d} t .\) 

From the boundary conditions, the first term is zero and we get
\(z (1-z)\left( \int _0^1 \frac{f(t)}{t-z}\, \operatorname{d}t\right)^\prime =-\int _{0}^1 f(t) \operatorname{d}t+\int _{0}^1 \frac{t(1-t) f^{\prime }(t)}{t-z} \operatorname{d}t .\) 

We return back to (REF ), and see that this is equivalent to prove that
\(\lim _{z \rightarrow 0}z^2 (1-z) \left( \int _0^1 \frac{f(t)}{t-z}\, \operatorname{d}t\right)^\prime = 0_N .\) 

This follows from the fact that the Stieltjes transform of \(z(1-z)f^\prime (z)\) , i.e.
\(\int _{0}^1 \frac{t(1-t) f^{\prime }(t)}{t-z} \operatorname{d}t ,\) 

is of the same type of \(f\) .
Then,
\(\lim _{z\rightarrow 0} z\left( \int _0^1 \frac{ t(1-t)f^\prime (t)}{t-z}\, \operatorname{d}t \right)=0_N, &&\lim _{z\rightarrow 1} (1-z)\left( \int _0^1 \frac{t(1-t)f^\prime (t)}{t-z} \, \operatorname{d}t\right)=0_N ,\) 

and (REF ) follows.
Now, as each entry of the matrix \(Q_{n}^{\mathsf {L}}(z)\)  is a Cauchy transform of certain function \(f\)  described previously, by using (REF ) and (REF ) we have that,
\(\left(Y_{n}^{\mathsf {L}}\right)^{\prime }(z)=\left[\begin{matrix}\operatorname{O}(1) & \operatorname{o}(\frac{1}{z^2}) \\\operatorname{O}(1) & \operatorname{o}(\frac{1}{z^2})\end{matrix}\right],&&\left(Y_{n}^{\mathsf {L}}(z)\right)^{-1}=\left[\begin{matrix}\operatorname{o}(\frac{1}{z}) & \operatorname{o}(\frac{1}{z}) \\\operatorname{O}(1) & \operatorname{O}(1)\end{matrix}\right],&& z \rightarrow 0\) 

and
\(\left(Y_{n}^{\mathsf {L}}\right)^{\prime }(z)=\left[\begin{matrix}\operatorname{O}(1) & \operatorname{o}(\frac{1}{(1-z)^2}) \\\operatorname{O}(1) & \operatorname{o}(\frac{1}{(1-z)^2})\end{matrix}\right], &&\left(Y_{n}^{\mathsf {L}}(z)\right)^{-1}=\left[\begin{matrix}\operatorname{o}(\frac{1}{1-z}) & \operatorname{o}(\frac{1}{1-z}) \\\operatorname{O}(1) & \operatorname{O}(1)\end{matrix}\right], &&z \rightarrow 1 .\) 

This implies that
\(\lim _{z \rightarrow 0}z^{2} \left(Y_{n}^{\mathsf {L}}\right)^{\prime }(z)\left(Y_{n}^{\mathsf {L}}\right)^{-1}& =\lim _{z \rightarrow 0} z^{2}\left[\begin{matrix}\operatorname{o}(\frac{1}{z})+ \operatorname{o}(\frac{1}{z^2}) &\operatorname{o}(\frac{1}{z})+\operatorname{o}(\frac{1}{z^2}) \\\operatorname{o}(\frac{1}{z^2})+\operatorname{o}(\frac{1}{z}) & \operatorname{o}(\frac{1}{z^2})+ \operatorname{o}(\frac{1}{z})\end{matrix}\right]\\ &=\lim _{z \rightarrow 0} z^{2}\left[\begin{matrix}\operatorname{o}(\frac{1}{z^2}) &\operatorname{o}(\frac{1}{z^2}) \\\operatorname{o}(\frac{1}{z^2}) &\operatorname{o}(\frac{1}{z^2})\end{matrix}\right]=0_{2 N} ,\) 

and
\(\lim _{z \rightarrow 1} (1-z)^{2}\left(Y_{n}^{\mathsf {L}}\right)^{\prime }(z)\left(Y_{n}^{\mathsf {L}}\right)^{-1}=\lim _{z \rightarrow 1} (1-z)^{2}\left[\begin{matrix}\operatorname{o}(\frac{1}{(1-z)^2}) &\operatorname{o}(\frac{1}{(1-z)^2}) \\\operatorname{o}(\frac{1}{(1-z)^2}) &\operatorname{o}(\frac{1}{(1-z)^2})\end{matrix}\right]= 0_{2 N} .\) 

Straightforward calculation and similar considerations lead us to
\(\lim _{z \rightarrow 0} z Y_{n}^{\mathsf {L}}(z)\left[\begin{matrix}h^{\mathsf {L}}(z) & 0_{N} \\0_{N} & -h^{\mathsf {R}}(z)\end{matrix}\right]\left(Y_{n}^{\mathsf {L}}(z)\right)^{-1} & =0_{2 N},\\\lim _{z \rightarrow 1} (1-z) Y_{n}^{\mathsf {L}}(z)\left[\begin{matrix}h^{\mathsf {L}}(z) & 0_{N} \\0_{N} & -h^{\mathsf {R}}(z)\end{matrix}\right]\left(Y_{n}^{\mathsf {L}}(z)\right)^{-1} & =0_{2 N} .\) 

Finally we arrive to
\(\lim _{z \rightarrow 0} z^{2} M_{n}^{\mathsf {L}}(z)=0_{2 N}&&\text{and}&&\lim _{z \rightarrow 1} (1-z)^{2} M_{n}^{\mathsf {L}}(z)=0_{2 N} .\) 

By analogous arguments we get the results for \(M_n^{\mathsf {R}}\) .

